Q96 / 160 · AP EAMCET 2014 · Mathematics

If $\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$, then $(A, B, C)$ equals to

  • A
    $(1,1,1)$
  • B
    $(1,1,0)$
  • Correct answer
    C
    $(1,0,1)$
  • D
    $(1,2,1)$

Solution

$\begin{aligned} & \text {Given, } \frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5} \\ & \therefore 2 x^3+x^2-5=(A x+B)\left(2 x^2+5\right) \\ & +(C x+1)\left(x^2-5\right) \\ & \Rightarrow 2 x^3+x^2-5=A x^3+5 A x+B x^2+5 B \\ & +C x^3-5 C x+x^2-5 \\ & \Rightarrow 2 x^3+x^2-5=x^3(A+C)+x^2(B+1) \\ & +x(5 A-5 C)+5 B-5 \end{aligned}$ On equating the coefficients of $x^3, x^2, x$ and constant, we get $\begin{aligned} & A+C=2, B+1=1,5 A-5 C=0 \\ & \text {and } \\ & 5 B-5=-5 \\ & B+1=1 \Rightarrow B=0 \\ & 5 A-5 C=0 \Rightarrow A=C \\ & A+C=2 \Rightarrow C+C=2 \\ & \Rightarrow \quad 2 C=2 \Rightarrow C=1 \end{aligned}$ Now, and $\begin{aligned} & \therefore & A & =C=1 \\ & \therefore & (A, B, C) & =(1,0,1)\end{aligned}$

Hence, the correct answer is option: 'C'.

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